Under what conditions will the ring homomorphism $\phi : R \to S$ satisfy the following results about prime and maximal ideals?

Question

Let $R$ and $S$ be rings and $\phi : R \to S$ be a ring homomorphism.

Here, I am considering that $R$ and $S$ don't necessarily have multiplicative identities.

MOTIVATION :

I know that the pre-image of a prime ideal is prime if $R$ and $S$ are commutative. But is this a necessary condition ? Similarly, I know that the pre-image of a maximal ideal need not be maximal even if $R$ and $S$ are commutative. But then, what are the sufficient conditions for this to hold ? And what are the sufficient conditions ?

QUESTION :

Basically, I am interested in knowing the necessary conditions and the sufficient conditions for the following to hold:

1. Image of a prime ideal is prime.
2. Pre-image of a prime ideal is prime.
3. Image of a maximal ideal is maximal.
4. Pre-image of a maximal ideal is maximal.

These conditions can be on the rings (e.g. commutativity, having unity, etc.) or on the homomorphism itself (e.g. injectivity, surjectivity, etc.)

P.S.

I know that this is a lot to ask! And some arguments might be trivial to see. But my aim to compare these situations with each other to get a better understanding of what's going on. Also, this might serve as a one-stop solution for anyone interested in any of these problems and trying to get some help from this website.

Any help/input shall be highly appreciated. Thanks in advance.

Answer 1

I can tell provide an answer for the case of commutative rings having unity.

Proposition 1: Image of a prime(respectively maximal) ideal containing the ideal $I:=\ker \phi$ of $R$ is prime(respectively maximal) in the subring $R':= Im \phi$ of $S$(show that is it indeed a subring!).(I'm not sure if this is true in the general case but I suspect not).

Proposition 2: The pre-image of a prime ideal is prime. This is NOT true in general for maximal ideals (see counterexample below).

Proposition 3: The pre-image of maximal ideal under a surjective ring homomorphism is maximal.

Proof of Proposition 1: Let $M$ be a prime(respectively maximal) ideal of $R$ containing $I$ and let , so that $\phi:R \twoheadrightarrow R'$ is surjective. Let $M':=\phi(M)$ be an ideal in $R'$(show that it is indeed an ideal!). We have the following chain of maps $$R \xrightarrow{\phi} R' \xrightarrow{\psi} R'/M'$$ where $\psi:R' \twoheadrightarrow R'/M'$ is the natural ring homomorphism(surjective as well). $\ker(\phi)=I, \ker(\psi)=M'=\phi(M) \implies \ker(\psi \circ \phi)= M \cup I=M$

So the First Isomorphism theorem on the composition map gives $R/M \cong R'/M'$

So, $I \subseteq M \subseteq R$ is prime(respectively maximal) implies $R/M \cong R'/M'$ is an integral domain(repectively field) implies $M'=\phi(M)$ is prime(respectively maximal) in $R'=Im(\phi)$.

Proof of Proposition 2: Let $P \subseteq S$ be a prime ideal. It is easy to show that $\phi^{-1}(P)$ is an ideal. Let $a,b \in R$ such t hat $ab \in \phi^{-1}(P) \implies \phi(a)\phi(b) \in P \implies \phi(a) \in P$ or $\phi(b) \in P \implies a \in \phi^{-1}(P)$ or $b \in \phi^{-1}(P)$. Hence $\phi^{-1}(P)$ is prime.

Counter-example for maximal ideals: $\phi:\mathbb{Z} \to \mathbb{Q}$ be the natural inclusion ring homomorphism. The Zero Ideal $I=(0)$ is a maximal ideal of $\mathbb{Q}$(since it is a field), but $\phi^{-1}((0))=(0)$ is clearly not a maximal ideal of $\mathbb{Z}$.

Proof of Proposition 3: This follows directly from the converse of Proposition 1(this holds too!, just reverse all steps in the proof) Try to show this yourself.